Challenging +1.8 This is a challenging M3 energy conservation problem requiring careful analysis of two elastic strings with different properties. Students must identify that maximum KE occurs when the net force is zero (equilibrium position), set up force balance equations to find this position, then apply energy conservation accounting for elastic PE in both strings and gravitational PE changes. The multi-step reasoning, coordinate geometry, and simultaneous consideration of two energy stores makes this significantly harder than average.
\includegraphics{figure_4}
\(A\) and \(C\) are two fixed points, \(1.5\) m apart, on a smooth horizontal plane. A light elastic string of natural length \(0.4\) m and modulus of elasticity \(20\) N has one end fixed to point \(A\) and the other end fixed to a particle \(B\). Another light elastic string of natural length \(0.6\) m and modulus of elasticity \(15\) N has one end fixed to point \(C\) and the other end fixed to the particle \(B\). The particle is released from rest when \(ABC\) forms a straight line and \(AB = 0.4\) m (see diagram).
Find the greatest kinetic energy of particle \(B\) in the subsequent motion. [7]
Initial energy = \(\frac{15 \times 0.5^2}{2 \times 0.6}\)
\(\frac{20x}{0.4} = \frac{15(0.5-x)}{0.6}\)
\(x = \frac{1}{6}\) (m)
Answer
Marks
Guidance
\(\frac{20x(\frac{1}{6})^2}{2 \times 0.4}\) or \(\frac{15(0.5-\frac{1}{6})^2}{2 \times 0.6}\) seen
B1, M1A1, A1, B1
3.125 (J) or \(\frac{25}{8}\). \(x\) measured from \(B\); other points possible. OR \(\frac{17}{30}\) from \(A\), or \(\frac{11}{60}\) from midpt. Allow their \(\frac{1}{6}\)
\(x = 0\) or \(\frac{1}{6}\) so max EPE when \(x = \frac{1}{6}\)
(M1A1), (A1)
B1 & B1M1A1 as above
Or by SHM, \(m\dot{x} = \frac{15(0.5-x)}{0.6} - \frac{20x}{0.4}\)
Answer
Marks
Guidance
\(m\ddot{x} = -75\left(x - \frac{1}{6}\right)\)
(M1A1), (A1)
Or other \(x\)
Amplitude, \(a = \frac{1}{6}\)
\(ω^2 = \frac{75}{m}\)
Max KE = \(\frac{1}{2}m\left(\frac{1}{6}\right)^2 \times \frac{75}{m}\)
Answer
Marks
\(\frac{25}{24}\) J (or 1.04)
(B1), (B1), (M1), (A1)
Initial energy = $\frac{15 \times 0.5^2}{2 \times 0.6}$
$\frac{20x}{0.4} = \frac{15(0.5-x)}{0.6}$
$x = \frac{1}{6}$ (m)
$\frac{20x(\frac{1}{6})^2}{2 \times 0.4}$ or $\frac{15(0.5-\frac{1}{6})^2}{2 \times 0.6}$ seen | B1, M1A1, A1, B1 | 3.125 (J) or $\frac{25}{8}$. $x$ measured from $B$; other points possible. OR $\frac{17}{30}$ from $A$, or $\frac{11}{60}$ from midpt. Allow their $\frac{1}{6}$ | Or 1/3 + 0.6 from C. Not their $\frac{17}{30}$ or $\frac{11}{60}$
KE = 3.125 $= \frac{20x(\frac{1}{6})^2}{2 \times 0.4} - \frac{15(0.5-\frac{1}{6})^2}{2 \times 0.6}$
$= 1.04167$ (J) | M1 | Allow their $\frac{1}{6}$ | $\frac{25}{8}, \frac{25}{36}, \frac{25}{18}$, $\frac{75}{72}$
OR KE = [3.125] $- \frac{20x^2}{2 \times 0.4} - \frac{15(0.5-x)^2}{2 \times 0.6}$
Differentiate OR complete the square | (M1), (M1) | All terms present. Allow slips | CTS $-37.5(\frac{1}{6} - x)^2 + \frac{37.5}{36}$ leads to max KE = $\frac{37.5}{36}$
Max KE when $x = \frac{1}{6}$ | (A1) | B1 & B1M1A1 as above
OR stationary when
$3.125 = \frac{20x^2}{2 \times 0.4} + \frac{15(0.5-x)^2}{2 \times 0.6}$
$x = 0$ or $\frac{1}{6}$ so max EPE when $x = \frac{1}{6}$ | (M1A1), (A1) | B1 & B1M1A1 as above
Or by SHM, $m\dot{x} = \frac{15(0.5-x)}{0.6} - \frac{20x}{0.4}$
$m\ddot{x} = -75\left(x - \frac{1}{6}\right)$ | (M1A1), (A1) | Or other $x$
Amplitude, $a = \frac{1}{6}$
$ω^2 = \frac{75}{m}$
Max KE = $\frac{1}{2}m\left(\frac{1}{6}\right)^2 \times \frac{75}{m}$
$\frac{25}{24}$ J (or 1.04) | (B1), (B1), (M1), (A1) |
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\includegraphics{figure_4}
$A$ and $C$ are two fixed points, $1.5$ m apart, on a smooth horizontal plane. A light elastic string of natural length $0.4$ m and modulus of elasticity $20$ N has one end fixed to point $A$ and the other end fixed to a particle $B$. Another light elastic string of natural length $0.6$ m and modulus of elasticity $15$ N has one end fixed to point $C$ and the other end fixed to the particle $B$. The particle is released from rest when $ABC$ forms a straight line and $AB = 0.4$ m (see diagram).
Find the greatest kinetic energy of particle $B$ in the subsequent motion. [7]
\hfill \mbox{\textit{OCR M3 2016 Q4 [7]}}